Let $E$ be a two-dimensional real normed space. In this paper we show that if
the unit circle of $E$ does not contain any line segment such that the distance
between its endpoints is greater than 1, then every transformation $\phi\colon
E\to E$ which preserves the unit distance is automatically an affine isometry.
In particular, this condition is satisfied when the norm is strictly convex.Comment: 8 pages, 3 figure

Gehér, György Pál

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György Pál Gehér

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Linear Algebra and its Applications

A contribution to the Aleksandrov conservative distance problem in two dimensions

Gehér, György

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A CONTRIBUTION TO THE ALEKSANDROVCONSERVATIVE DISTANCE PROBLEM IN TWODIMENSIONSGYO¨RGY PA´L GEHE´RAbstract. Let E be a two-dimensional real normed space. In thispaper we show that if the unit circle of E does not contain any linesegment such that the distance between its endpoints is greater than 1,then every transformation φ : E → E which preserves the unit distance isautomatically an affine isometry. In particular, this condition is satisfiedwhen the norm is strictly convex.1. IntroductionIn 1953 F. S. Beckman and D. A. Quarles characterized isometries ofn-dimensional Euclidean spaces under a surprisingly mild condition whenn ≥ 2 (see [2] or [4, 9] for alternative proofs). Namely, they managed toshow that every transformation φ : Rn → Rn which preservers unit Euclideandistance in one direction is an (affine) isometry. They also noted that on Ror on an infinite dimensional, real Hilbert space the same conclusion fails.Many mathematicians have been trying to generalize this beautiful the-orem. The problem of characterizing those finite dimensional real normedspaces E such that every transformation φ : E → E which preserves the unitdistance in one direction is an isometry was raised, in this general form, by A.D. Aleksandrov and hence it is called the Aleksandrov conservative distanceproblem (see [1]). In the literature these spaces are also called Beckman-Quarles type spaces. As far as we know, the original version of Aleksandrovproblem was solved only for a few concrete normed spaces (see [22] concern-ing p-norms, and [13] where the norm is not strictly convex), all of themare two-dimensional. Some general results are known for modified versions,for instance in [5] W. Benz and H. Berens investigated the case when thetransformation preservers distance 1 and n for some n ∈ N, n > 1. We alsomention the paper [18] of T. M. Rassias and P. Sˇemrl where they assumedthat φ is onto and it preserves distance 1 in both directions. They showedthat in this case φ is not very far from being an isometry. Several other2010 Mathematics Subject Classification. Primary: 46B04, 46B20, 52A10. Secondary:51M05, 52C25.Key words and phrases. Aleksandrov conservative distance problem, Beckman-Quarlestype space, isometry, unit distance preserving mapping.12 GYO¨RGY PA´L GEHE´Rresults are known which are connected to the Aleksandrov problem. Thereader can find a number of them in the References.The original version remained unsolved even for the very special casewhen dimE = 2 and the norm is strictly convex. Here we present a unifiedapproach which solves the Aleksandrov problem in two dimensions for amuch larger class of norms, which we will call URTC-norms. Let us pointout that the naive conjecture that every at least two but finite dimensionalnormed space is a Beckman-Quarles type space is false. However, as far aswe know, counterexamples are only known in the simple case when the unitball of the norm is a linear image of a cube (see [17]).2. Auxiliary definitions and statement of the main resultSince we will consider only two-dimensional normed spaces over R, we caninvestigate R2 endowed with a norm ‖ · ‖. We say that the norm is strictlyconvex, if its sphere S does not contain any non-degenerated line segment.If three points a, b, c ∈ R2 satisfy d = ‖a− b‖ = ‖b− c‖ = ‖c− a‖ for somed > 0, then these points are said to be in a regular d-position. We introducethe following notion.Definition 1. We call ‖ · ‖ a URTC-norm (unique regular triangle con-structibility) if for every a, b ∈ R2, ‖a− b‖ = 1 the equation system{‖a− x‖ = 1‖b− x‖ = 1 (1)is satisfied exactly for two points x ∈ R2.Since the function f(x) := ‖b − x‖ is continuous on a + S, f(b) = 0and f(2a − b) = 2, the existence of such an x which fulfilles (1) is trivial.Obviously, if x satisfies (1), then a+ b− x 6= x fulfilles it as well.By translation, we may have assumed that a = 0, and by multiplyingwith a non-zero scalar, we may have replaced 1 by any d > 0 in Definition1. We note that for the `∞ norm, one can find two points a, b ∈ R2 with‖a − b‖ = 1 such that (1) holds for infinitely many points x ∈ R2. We willprovide a useful characterization of URTC-norms in Lemma 1.Our main theorem, which reads as follows and will be proven in Section3, provides an affirmative answer for the Aleksandrov conservative distanceproblem for URTC-noms.Theorem 1. Let ‖ · ‖ be a URTC-norm on R2, and let us consider anarbitrary transformation φ : R2 → R2 such thatx, y ∈ R2, ‖x− y‖ = 1 =⇒ ‖φ(x)− φ(y)‖ = 1.Then φ is an affine isometry.We will need several lemmas before proving Theorem 1. We note thatTheorem 1 can be considered as a Mazur-Ulam type result in two dimensions(see [14, 7]). Let us point out that quite the same proof works for the caseTHE ALEKSANDROV PROBLEM IN TWO DIMENSIONS 3if we consider two different URTC-norms on the initial and final spaces.However, dealing with the above version makes notations much simpler.Furthermore, by affinity, the modified version of our main theorem saysin many cases (in fact when the unit circles of these norms are not linearimages of each other) that no transformation φ exists which preserves theunit distance.3. Proof of the main resultWe begin with a characterization of URTC-norms. The symbols [a, b] and`(a, b) will denote the line segment {a + t(b − a) : 0 ≤ t ≤ 1} and the line{a+ t(b− a) : t ∈ R}, respectively.Lemma 1. The following conditions are equivalent for any norm ‖ · ‖ onR2:(i) ‖ · ‖ is not a URTC-norm,(ii) two points c, d ∈ S, ‖c− d‖ > 1 exist such that [c, d] ⊆ S.In particular, every strictly convex norm is a URTC-norm.Proof. (ii)=⇒(i): Set a = 0 and b = 1‖d−c‖(d − c) ∈ S. Then every x ∈[c+ b, d] satisfies (1). Therefore the norm cannot be URTC.(i)=⇒(ii): Since the norm does not have the URTC property, there existsa point b ∈ S such that at least three different solutions can be given whichsatisfy (1) with a = 0. Clearly, none of them can lie on `(0, b). Thereforeat least two of them, x and y, lies on the same open side of `(0, b). First,we show that x ∈ `(y, y − b). Assuming the contrary, we can suppose,without loss of generality, that x lies between `(0, b) and `(y, y− b). Clearlyconv(b, y, y−b,−b)∩S ⊆ [b, y]∪[y, y−b]∪[y−b,−b], but since ‖x−b‖ = ‖y−b‖holds, we get x /∈ [b, y]. On the one hand, if x lies on the opposite closed sideof `(−b, y−b) than b, then x has to be in the interior of conv(0, y, x−b). Sincey, x, x−b ∈ S, this is impossible. On the other hand, if x lies on the oppositeside of `(b, y) than 0, then x− b has to be in the interior of conv(0, x, y− b)which is again a contradiction. Therefore, indeed x ∈ `(y, y− b) is satisfied.Now, we may suppose that x− y = ‖x− y‖b. Since y− b, y, x are distinctcollinear points of the unit sphere, we easily obtain that [y − b, x] ⊆ S and‖(y − b)− x‖ > 1, which completes the proof of this part. The shorter closed and open arcs of S between two non-antipodal pointsb0, b1 will be denoted by arc(b0, b1) and arc◦(b0, b1), respectively. We providesome basic properties of URTC-norms in the following lemma.Lemma 2. (i) Let ‖ · ‖ be a URTC-norm. If a, b, c, d ∈ S such that‖a−d‖ = ‖b−c‖ = 1 and (a−d, a, b, c, d, d−a) is positively oriented,then we have a = b and c = d.(ii) Let ‖ · ‖ be an arbitrary norm. Suppose that we have a, b ∈ R2, 0 <γ := ‖a− b‖ and 0 ≤ α, β. If |β − γ| ≤ α ≤ β + γ is satisfied, thenthere exists a point c ∈ R2 which fulfilles ‖a−c‖ = β and ‖b−c‖ = α.4 GYO¨RGY PA´L GEHE´RProof. (i): Clearly we have arc◦(a, d) ⊆ conv(a, d, a+ d), and by the URTCproperty arc◦(a, d)∩ ([a, a+d]∪ [d, a+d]) = ∅ is valid. Elementary observa-tions show that if the Euclidean distance between b and `(a, d) is less thanor equal to the distance of c and `(a, d), then c− b ∈ conv(0, d− a, d) holds.This implies c− b ∈ [d− a, d]. Thus we obtain b = a and d ∈ [d− a, c], andhence [d− a, c] ⊆ S is satisfied. Since our norm is URTC, we obtain c = d.The other case can be shown similarly.(ii): We consider the continuous function F : β ·S+a→ R, F (z) = ‖b−z‖.Since we have F(a+ βγ (b− a))= |β − γ| ≤ α and F(a− βγ (b− a))=β+γ ≥ α, we conclude the existence of a point c ∈ R2 such that ‖a−c‖ = βand ‖b− c‖ = α holds. We define the functions f, g : S → S, z 7→ f(z) such that ‖z − f(z)‖ = 1,‖z − g(z)‖ = 1, and (0, z, f(z)), (0, g(z), z) are positively oriented. By theURTC property, f and g are well-defined, moreover, we clearly have g−1 =f , and hence f and g are bijective. We proceed with showing that f iscontinuous.Lemma 3. The function f is continuous.Proof. We assume indirectly that f is not continuous, and without loss ofgenerality we may suppose that b1 ∈ S is a point of discontinuity. We setb0 = g(b1) and b2 = f(b1).A quite straightforward application of Lemma 2 gives the followingmonotonicity property of f : if z ∈ arc◦(z0, f(z0)), then we have f(z) ∈arc◦(f(z0), f(f(z0))). By this monotonicity property we obtain that thereare two points b2−ε ∈ arc◦(b1, b2) and b2+ε ∈ arc◦(b2,−b0) such that ei-ther f(S) ∩ arc◦(b2−ε, b2) = ∅, or f(S) ∩ arc◦(b2, b2+ε) = ∅. Both of themcontradicts to the bijectivity of f . Let d > 0. We call the 7-tupple (a, b1, b2, b3, c1, c2, c3) ∈ (R2)7 a d-probeifd = ‖a− b1‖ = ‖a− b2‖ = ‖b1 − b2‖ = ‖b1 − b3‖ = ‖b2 − b3‖= ‖a− c1‖ = ‖a− c2‖ = ‖c1 − c2‖ = ‖c1 − c3‖ = ‖c2 − c3‖ = ‖b3 − c3‖(2)holds ([16, 9], see also Figure 1). By the following lemma, any three pointswhich are in a regular d-position can be extended to a d-probe.Lemma 4. Let ‖ · ‖ be a URTC-norm, and let b1, b2 ∈ d · S such that‖b1 − b2‖ = d. Then the 3-tupple (0, b1, b2) can be extended to a d-probe(0, b1, b2, b3, c1, c2, c3), where we necessarily have b3 = b1 + b2. Moreover,0 6= b3 holds.Proof. We can assume without loss of generality that (0, b1, b2) is positivelyoriented. Let us defineh : d · S → R, z 7→∥∥∥∥12(z + f(z))− 12(b1 + b2)∥∥∥∥ ,THE ALEKSANDROV PROBLEM IN TWO DIMENSIONS 5Figure 1. A d-probe. An edge represents that the distancebetween the endpoints is exactly d.which is trivially continuous and h(b1) = 0. Moreover, by the triangleinequality we obtainh(−b1) =∥∥∥∥12(−b1 − b2)− 12(b1 + b2)∥∥∥∥ = ‖b1 + b2‖= ‖2b2 − (b2 − b1)‖ ≥ ‖2b2‖ − ‖(b2 − b1)‖ = d,which immediately implies the existence of a c1 ∈ d · S such that h(c1) = d2 .We define b3 = b1 + b2, c2 = f(c1) and c3 = c1 + c2. The ordered 7-tupple(0, b1, b2, b3, c1, c2, c3) is trivially a d-probe.In the above construction we chose b3 to be b1 + b2. Now, we show thatthis is the only choice. Since the norm is URTC, the first line of (2) (witha = 0) implies b3 ∈ {0, b1 + b2}. Suppose that 0 = b3 happens. Since‖b3 − c3‖ = d, we obtain c3 = c1 + c2. But from c1 − c2, c1, c1 + c2 ∈ d · Swe conclude [c1 − c2, c1 + c2] ⊆ d · S, which clearly contradicts the URTCproperty. Therefore we indeed have b3 = b1 + b2. Now, we are in the position to present the proof of the main result of thispaper.Proof of Theorem 1. Suppose that φ preserves distance d > 0. Letb0, b1, b2 ∈ R2 be arbitrary three points which are in a regular d-position. Letus consider a d-probe (b0, b1, b2, b3, c1, c2, c3) ∈ (R2)7 which exists by Lemma4 and where b3 = b1+b2−b0. Clearly, the 7-tupple (φ(b0), φ(b1), φ(b2), φ(b3),φ(c1), φ(c2), φ(c3)) ∈ (R2)7 is a d-probe as well, and therefore φ(b3) =φ(b1 + b2 − b0) = φ(b1) + φ(b2) − φ(b0) is satisfied. Set b4 = 2b2 − b0.Considering b1, b2, b3 instead of b0, b1, b2, by the previous observations weconclude φ(b4) = φ(2b2 − b0) = 2φ(b2) − φ(b0). This immediately impliesthat distance 2d is also preserved, moreover, when b0, b2, b4 are collinear suchthat d = ‖b0 − b2‖ = ‖b2 − b4‖ and 2d = ‖b0 − b4‖, then the same hold fortheir images.6 GYO¨RGY PA´L GEHE´RFigure 2Iterating the above method we can easily prove also the following state-ment: distance nd is also preserved for every n ∈ N, furthermore, whena, b, c ∈ R2 are collinear such that d = ‖a − b‖, (n − 1)d = ‖b − c‖ andnd = ‖a − c‖, then the same is valid for their images. In particular, φpreserves distance n for all n ∈ N (see Figure 2).Figure 3Next, we show that assuming distance d is preserved implies that distancedn is preserved as well for every n ∈ N. Let a, b ∈ R2, ‖a − b‖ = dn . ByLemma 2, there exists a point c ∈ R2 such that ‖a− c‖ = ‖b− c‖ = d. Sete = c+n(a−c) and f = c+n(b−c). Obviously, we have ‖a−e‖ = ‖b−f‖ =(n − 1)d and ‖e − f‖ = d. Therefore we have d = ‖φ(a) − φ(c)‖ = ‖φ(b) −φ(c)‖ = ‖φ(e)−φ(f)‖, (n− 1)d = ‖φ(a)−φ(e)‖ = ‖φ(b)−φ(f)‖, moreover,φ(c), φ(a), φ(e) are collinear and φ(c), φ(b), φ(f) are collinear. This impliesφ(a)−φ(b) = 1n(φ(e)−φ(f)), and thus ‖φ(a)−φ(b)‖ = 1n‖φ(e)−φ(f)‖ = dn .(See Figure 3).By the above observations, we immediately obtain that φ preserves allrational distances. Let a, b ∈ R2 be two arbitrary different points. For every0 < ε < ‖a−b‖3 we can find p, q ∈ Q such that 0 < q < ε and p − q <‖a− b‖ < p+q. By Lemma 2 we can find such a point c ∈ R2 which satisfies‖a− c‖ = p, ‖b− c‖ = q. Since rational distances are preserved by φ, we get‖φ(a)− φ(c)‖ = p, ‖φ(b)− φ(c)‖ = q, and by the triangle inequalityp− ε < p− q ≤ ‖φ(a)− φ(b)‖ ≤ p+ q < p+ ε.Since this holds for every 0 < ε < ‖a−b‖3 , we conclude ‖φ(a)−φ(b)‖ = ‖a−b‖,which means that φ is indeed an isometry.Since isometries are continuous, affinity of φ follows from the preservationsof midpoints which was pointed out before. This completes the proof. THE ALEKSANDROV PROBLEM IN TWO DIMENSIONS 7Several norms on R2 do not have the URTC property. It is not clear whatthe answer is for the Aleksandrov conservative distance problem for thesenorms. Those techniques which were presented here do not work for theseclass of norms. We left this question as a challenging open problem.Acknowledgements.The author was supported by the ”Lendu¨let” Program (LP2012-46/2012)of the Hungarian Academy of Sciences.References[1] A. D. Aleksandrov, Mapping of families of sets, Soviet Math. Dokl. 11 (1970), 116–120.[2] F. S. Beckman and D. A. Quarles, On isometries of Euclidean spaces, Proc. Amer.Math. Soc. 4 (1953), 810–815.[3] W. Benz, An elementary proof of the theorem of Beckman and Quarles, Elem.Math. 42 (1987), 4–9.[4] W. Benz, Real Geometries, Bibliographisches Institut, Mannheim, 1994.[5] W. Benz and H. Berens, A contribution to a theorem of Ulam and Mazur, Aequa-tiones Math. 34 (1987), 61–63.[6] H.-Y. Chu, S. K. Choi and D. S. Kang, Mappings of conservative distances in linearn-normed spaces, Nonlinear Anal. 70 (2009), 1168–1174.[7] R. J. Fleming and J. E. Jamison Isometries on Banach Spaces: Function Spaces,Chapman & Hall/CRC Monographs and Surveys in Pure and Applied Mathematics,129.[8] W. Hua, Geradenbildungen, die den Fla¨cheninhalt 1 von Dreiecken erhalten, Ham-burger Beitra¨ge zur Math. aus dem Math. Seminar, Heft 22 (1993).[9] R. Juha´sz, Another Proof of the BeckmanQuarles Theorem, to appear.[10] S.-M. Jung, Inequalities for distances between points and distance preserving map-pings, Nonlinear Anal. 62 (2005), 675–681.[11] A. Kubzdela, Isometries, Mazur-Ulam theorem and Aleksandrov problem for non-Archimedean normed spaces, Nonlinear Anal. 75 (2012), 2060–2068.[12] J. A. Lester, Martin’s theorem for Euclidean n-space and a generalization to theperimeter case, J. Geom. 27 (1986), 29–35.[13] J. M. Ling, A normed space with the Beckman-Quarles property, Beitra¨ge AlgebraGeom. 48 (2007), 131–139.[14] S. Mazur and S. Ulam, Sur les transformationes isome´triques d’espaces vectorielsnorme´s, C. R. Acad. Sci. Paris 194, 946–948.[15] M. S. Moslehian and G. Sadeghi, A Mazur-Ulam theorem in non-Archimedeannormed spaces Nonlinear Anal. 69 (2008), 3405–3408.[16] D. E. Raiskii, The realization of all distances in a decomposition of the space Rninto n+ 1 parts, Math. Notes 7 (1970), 194–196.[17] T. M. Rassias, On the A. D. Aleksandrov problem of conservative distances andthe Mazur-Ulam theorem, Proceedings of the Third World Congress of NonlinearAnalysts, Part 4 (Catania, 2000). Nonlinear Anal. 47 (2001), 2597–2608.[18] T. M. Rassias and P. Sˇemrl, On the Mazur-Ulam theorem and the Aleksandrovproblem for unit distance preserving mappings, Proc. Amer. Math. Soc. 118 (1993),919–925.[19] L. Tan and S. Xiang, On the Aleksandrov-Rassias problem and the Hyers-Ulam-Rassias stability problem, Banach J. Math. Anal. 1 (2007), 11–22.[20] A. Tyszka, Beckman-Quarles type theorems for mapping from Rn to Cn, Aequa-tiones Math. 67 (2004), 225–235.8 GYO¨RGY PA´L GEHE´R[21] J. Va¨isa¨la¨, A proof of the Mazur-Ulam theorem, Amer. Math. Monthly 110 (2003),633–635.[22] R. WeiYun, On the Aleksandrov problem for two-dimensional `p space, NonlinearFunct. Anal. Appl. 11 (2006), 755–758.Bolyai Institute, University of Szeged, H-6720 Szeged, Aradi ve´rtanu´k tere1, HungaryMTA-DE ”Lendu¨let” Functional Analysis Research Group, Institute ofMathematics, University of Debrecen, H-4010 Debrecen, P.O. Box 12, Hun-garyE-mail address: gehergy@math.u-szeged.hu; gehergyuri@gmail.comURL: http://www.math.u-szeged.hu/~gehergy/

A contribution to the Aleksandrov conservative distance problem in two
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